Use $1+\cos (2x)=2{\cos }^{2}x$, so $\sqrt{1+\cos (2x)}=\sqrt{2}|\cos x|$. The equation becomes

$|\cos x|={\tan }^{-1}(\tan x)$.

Recall ${\tan }^{-1}(\tan x)$ is a sawtooth: it equals $x-k\pi$ on each interval $\left(k\pi -\frac{\pi }{2},k\pi +\frac{\pi }{2}\right)$ and is undefined at odd multiples of $\frac{\pi }{2}$.

Sketching $y=|\cos x|$ (non-negative, periodic, hump shape) against $y={\tan }^{-1}(\tan x)$ (line segments of slope $1$) on each of the three open intervals and counting intersections:

On $\left(-\frac{3\pi }{2},-\frac{\pi }{2}\right)$: the sawtooth piece is the line $y=x+\pi$, which is negative in part of the interval; intersection with $|\cos x|\ge 0$ occurs once.

On $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$: here ${\tan }^{-1}(\tan x)=x$ and $|\cos x|=\cos x$; for $x<0$ the RHS is negative while LHS is non-negative, so the only solution is in $[0,\pi /2)$, exactly one intersection.

On $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$: the sawtooth piece is $y=x-\pi$, giving one intersection with the cosine hump.

Total number of solutions $=3$.